How do you integrate #int x^2cos(3x)# by integration by parts method?

Question

12907 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-03T10:13:23

#1/27(9x^2-2)sin3x+(2x)/9cos3x+C#

#I=intuv'dx=uv-intvu'dx#

#intx^2cos3xdx#

#u=x^2=>u'=2x#

#v'=cos3x=>v=1/3sin3x#

#:.I_1=x^2/3sin3x-int[(2x)/3sin3x]dx#

#:.I_1=x^2/3sin3x-2/3color(blue)(int[xsin3x]dx)#

we now have to use IBP on the integral highlighted in blue

#I_2=(int[xsin3x]dx)#

#u=x=>u'=1#

#v'=sin3x=>v=-1/3cos3x#

#I_2=(-x/3cos3x-int-1/3cos3xdx)#

#I_2=-x/3cos3x+1/9sin3x#

substituting back into #I_1#

#:.I_1=x^2/3sin3x-2/3(-x/3cos3x+1/9sin3x)+C#

#:.I_1=x^2/3sin3x+(2x)/9cos3x-2/27sin3x+C#

#I_1=(x^2/3-2/27)sin3x+(2x)/9cos3x+C#

#I_1=1/27(9x^2-2)sin3x+(2x)/9cos3x+C#